# What is the vertex form of y= 8x^2+17x+1 ?

Jan 26, 2016

$y = 8 {\left(x + \frac{17}{16}\right)}^{2} - \frac{257}{32}$

#### Explanation:

The vertex form of the trinomial is; $y = a {\left(x - h\right)}^{2} + k$

where (h , k ) are the coordinates of the vertex.

the x-coordinate of the vertex is x $= - \frac{b}{2 a}$

[from $8 {x}^{2} + 17 x + 1$

a = 8 , b = 17 and c = 1 ]

so x-coord$= - \frac{17}{16}$

and y-coord $= 8 \times {\left(- \frac{17}{16}\right)}^{2} + 17 \times \left(- \frac{17}{16}\right) + 1$

$= \cancel{8} \times \frac{289}{\cancel{256}} - \frac{289}{16} + 1$

$= \frac{289}{32} - \frac{578}{32} + \frac{32}{32} = - \frac{257}{32}$

Require a point to find a: if x = 0 then y=1 ie (0,1)

and so : 1= a${\left(\frac{17}{16}\right)}^{2} - \frac{257}{32} = \frac{289 a}{256} - \frac{257}{32}$

hence $a = \frac{256 + 2056}{289} = 8$

equation is : $y = 8 {\left(x + \frac{17}{16}\right)}^{2} - \frac{257}{32}$